The problem asks to factor and simplify the following trigonometric expression:

$10 \cos^2{x} + \cos{x} - 11$

This looks like a quadratic equation in terms of $\cos{x}$. Let's solve it step by step.

Step 1: Substitute $u = \cos{x}$

This makes the equation:

$10u^2 + u - 11 = 0$

Step 2: Solve the quadratic equation

Use the quadratic formula:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 10$, $b = 1$, and $c = -11$. Substituting these values:

$u = \frac{-1 \pm \sqrt{1^2 - 4(10)(-11)}}{2(10)}$

$u = \frac{-1 \pm \sqrt{1 + 440}}{20} = \frac{-1 \pm \sqrt{441}}{20} = \frac{-1 \pm 21}{20}$

This gives two solutions:

1. $u = \frac{-1 + 21}{20} = \frac{20}{20} = 1$
2. $u = \frac{-1 - 21}{20} = \frac{-22}{20} = -\frac{11}{10}$

Step 3: Back-substitute $u = \cos{x}$

• For $\cos{x} = 1$, the solution is:

  x = 0 + 2k\pi \quad \text{(where k is any integer)}

• For $\cos{x} = -\frac{11}{10}$, this is impossible because $\cos{x}$ is bounded by $-1 \leq \cos{x} \leq 1$.

Step 4: Factor the original expression

Using the valid solution $\cos{x} = 1$, the factorization of the quadratic equation corresponds to:

$10(\cos{x} - 1)(\cos{x} + \frac{11}{10}) = 0$

However, since $\cos{x} = -\frac{11}{10}$ is not a valid solution, the only valid factorization relevant to real solutions is:

$10(\cos{x} - 1) = 0$

Final Answer:

The solution to the equation is:

$\boxed{\cos{x} = 1}$

Thus, $x = 0 + 2k\pi$, where $k$ is any integer.

Do you need further clarification or have any questions?

Here are 5 related questions for deeper understanding:

1. How can we solve other trigonometric equations with quadratic forms?
2. What are the bounds of $\cos{x}$ and $\sin{x}$, and why do they matter?
3. What is the difference between factoring and solving a trigonometric equation?
4. How does the quadratic formula work in more complex trigonometric contexts?
5. Can other substitution methods help with trigonometric equations?

Tip: Always check the domain constraints when dealing with trigonometric functions to avoid extraneous solutions!